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3 years ago

Round to the nearest hundred thousand 894

Mathematics

1 answer:

Yuliya22 [10]3 years ago

8 0

Answer:

Step-by-step explanation:

please simplify i dont understand this is a hundred not a hundred thousand

if you mean nearest hundred the answer is 900

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a nunber is multiplied by 2. then that product is multiplied by 5. write to explain how you know what digit is in the ones place

siniylev [52]

Any number multiplied by 2 will give u an even number.
2 x 3 = 6
2 x 4 = 8
2 x 5 = 10

and any even number multiplied by 5 will end in 0
5 x 2 = 10
5 x 4 = 20
5 x 6 = 30

so the number in the ones place will be 0

6 0

3 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

B = bugs are present in a computer program.

D = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago

A bag contains 4 red marbles and 5 blue marbles. What is the probability of randomly selecting a blue marble and then, without r

vovangra [49]

1, A and D are the same thing But I would go with C. Hope this Helps!

3 0

3 years ago

1/4 + 1/2 + x = −3/4

Ostrovityanka [42]

1/4 + 1/2 + x = −3/4

get a common denominator 1/4

1/4 +2/4 +x = -3/4

combine like terms

3/4 +x =-3/4

subtract 3/4 from each side

x = -3/4 -3/4

x = -6/4

divide top and bottom by 2

x = -3/2

x = -1 1/2

Answer : x = -1 1/2

7 0

3 years ago

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Billy’s mother had five children. The first was named Lala, the second was named Lele, the third was named Lili, the fourth was

pantera1 [17]

Answer:

Billy

Step-by-step explanation:

come on really?

3 0

3 years ago

Read 2 more answers